It might help to know what level of math this is, what textbook, etc. There are so many processes in algebra that it might help us narrow it down if we were given more of the context.

Samantha R. | Math, History, Theatre, Test Prep, Study SkillsMath, History, Theatre, Test Prep, Study...

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1. Simplify

[(x^{2}+2x-3)]/[(x+3)(x-3)]=10+y^{2}+2y-4y+8-2y^{2}
The right side requires simplification in that we need to combine like terms.
[(x^{2}+2x-3)]/[(x+3)(x-3)]=10 + y^{2 }+ 2y - 4y +8 - 2y^{2}
[(x^{2}+2x-3)]/[(x+3)(x-3)]=(-2y^{2}+y^{2})+(2y-4y)+(10+8)
[(x^{2}+2x-3)]/[(x+3)(x-3)]= - y^{2 }- 4y + 18
It is now simplified.

2. Factor

The numerator (x^{2} + 2x - 3) of the left side can be factored...
[(x+3)(x-1)]/[(x+3)(x-3)]=-y^{2}-4y+18
[(x+3)(x-1)]/[(x+3)(x-3)]=-y^{2}-4y+18 [The (x+3) on the numerator and the denominator cancel each other out.]
(x-1)/(x-3)=-y^{2}-4y+18
It is now factored.

3. Evaluate

We can evaluate the fact that in its original form, the domain can include all integers except -3 and 3.
In its current form, the domain can include all integers except 3.

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